Calculus 2-A10 - Calculus II Calculus: Early Transcendental...

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Calculus II Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0-618-60624-6). Written Assignment 10 Complete the following textbook exercises for sections 13.3, 13.4, 13.5, and 13.6, and submit them to your mentor for correction and grading. Show all calculations. SECTION 13.3 Exercises Do exercises 10, 14, 18, 34, 52, 62, and 68 on pages 912–913 of the textbook. In Exercises 5– 28, find both first partial derivatives. 2z/2x = 2/2x(y 3 – 4xy 2 – 1) =-4y 2 2z/2x = 2/2y(y 3 – 4xy 2 – 1) = 3y 2 - 8xy 2z/2x = 2/2x(ln√xy) = (1/√xy)( 2/2x) (xy) 1/2 = (1/√xy) (1/2√xy)(y) = y/2xy = 1/(2x) 2z/2x = 2/2y(ln√xy) = (1/√xy)( 2/2y) (xy) 1/2 = (1/√xy) (1/2√xy)(x) = x/2xy = 1/(2y) 2z/2x = 2/2x(xy/x 2 + y 2 ) = (x 2 + y 2 )(y) – xy(2x)/ (x 2 + y 2 ) 2 = x 2 y+ y 3 - 2x 2 y /(x 2 + y 2 ) 2 = -y((x 2 + y 2 )/ (x 2 + y 2 ) 2 2z/2x = 2/2y(xy/x 2 + y 2 ) = (x 2 + y 2 )(x) – xy(2y)/ (x 2 + y 2 ) 2 = x 3 +xy 2 - 2xy 2 /(x 2 + y 2 ) 2 = -x(y 2 - x 2 )/(x 2 + y 2 ) 2 = x((x 2 + y 2 )/ (x 2 + y 2 ) 2
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In Exercises 33– 36, evaluate and at the given point. ƒx = ∂/∂x(arc cos xy) =1/√(1 – (xy)) 2 * ∂/∂x(xy) =-y/√(1 – (x 2 y 2 )) ƒy = ∂/∂y(arc cos xy) =-1/√(1 – (xy)) 2 * ∂/∂y(xy) =-x/√(1 – (x 2 y 2 )) at (1,1) ƒx = -(1)/√(1 – 1) = -∞ ƒy = (-1)/√(1 – 1) = -∞ Undefined In Exercises 51– 56, find the first partial derivatives with respect to x, y, and z. ∂w/∂x = ∂/∂x (3xz/x + y) = 3z [(x + y)(1) – x(1)/x + y) 2 ] = 3y 2 /(x + y) 2 = 3xz/(x + y) = 3xz * -1/(x + y) 2 * ∂/∂y (x + y) = -3xy/(x + y) 2 ∂w/∂z = ∂/∂z (3xz/x + y) = 3x [(x + y) * ∂/∂z (z) = 3x/(x + y) In Exercises 61– 68, find the four second partial derivatives. Observe that the second mixed partials are equal. ∂z/∂x = 4x
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Calculus 2-A10 - Calculus II Calculus: Early Transcendental...

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